Probability density expansion

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 1K views
Killtech
Messages
344
Reaction score
38
This is probably a stupid question but i don't want to make a stupid mistake here, so i thought better ask: I'm starting with the simple free Schrödinger Equation ##V(x)=0## (can be 1 dim) and an initial condition where the wave function is somehow constrained to be entirely localized around a compact set (let it be a sphere) ##S## around ##x=0## and let ##\Psi(x,0)=0## everywhere else. Just like a Gaussian wave package this wave function should disperse over time.

What's the easiest way to calculate ##\rho(d,t)=|\Psi(d,t)|^2## for an arbitrarily distant point ##d## outside ##S##?

I don't think i an can assume ##\Psi## to initially take the form of an indicator function since it's not differentiable around the edges and right now no other function with compact support comes into my mind that is easy to decompose into ##|p>## states for that matter. something like ##exp(\frac {1} {x^2-1})## in ##[-1; +1]## doesn't seem to be particularly friendly with Fourier transform.
 
  • Like
Likes   Reactions: Dale
Physics news on Phys.org
Killtech said:
This is probably a stupid question but i don't want to make a stupid mistake here, so i thought better ask: I'm starting with the simple free Schrödinger Equation ##V(x)=0## (can be 1 dim) and an initial condition where the wave function is somehow constrained to be entirely localized around a compact set (let it be a sphere) ##S## around ##x=0## and let ##\Psi(x,0)=0## everywhere else. Just like a Gaussian wave package this wave function should disperse over time.

What's the easiest way to calculate ##\rho(d,t)=|\Psi(d,t)|^2## for an arbitrarily distant point ##d## outside ##S##?

I don't think i an can assume ##\Psi## to initially take the form of an indicator function since it's not differentiable around the edges and right now no other function with compact support comes into my mind that is easy to decompose into ##|p>## states for that matter. something like ##exp(\frac {1} {x^2-1})## in ##[-1; +1]## doesn't seem to be particularly friendly with Fourier transform.

If you have a free particle with an initial wave-function that is known and is zero outside ##[-1, 1]##, then technically you express that initial wave-function as a distribution of free-particle states and apply time evolution. In general, that will get messy I imagine.
 
PeroK said:
If you have a free particle with an initial wave-function that is known and is zero outside ##[-1, 1]##, then technically you express that initial wave-function as a distribution of free-particle states and apply time evolution. In general, that will get messy I imagine.
that much i know, hence the "What's the easiest way" in my question ^^. Looking for a Fourier friendly initial function to have the least mess possible, especially when transforming the time developed thing back.
 
Killtech said:
that much i know, hence the "What's the easiest way" in my question ^^. Looking for a Fourier friendly initial function to have the least mess possible, especially when transforming the time developed thing back.

The classical example is the initial Gaussian. There are not many, I suspect, that result in a closed-form solution.
 
PeroK said:
The classical example is the initial Gaussian. There are not many, I suspect, that result in a closed-form solution.
I know, but the Gaussian does not have compact support but instead already fills the entirety of space right from the start. I could take the Fourier transform of an indicator function but not sure if that is a valid case given that its initial state violates Schrödinger by not being differentiable at the edges.